Consider Q to be the product (p-1)!. Then consider the product a.2a.3a.4a...(p-1)a. These are the same items rearranged, because a has an inverse.
But the product is also a^(p-1).(p-1)! by commutativity. So a^(p-1)=1 (mod p).
There's a clever idea in there, to consider prod(ka) for fixed a and k=1..(p-1). Not finding the clever idea means you won't prove it. Finding the clever idea is tricky. Once you know the clever idea the proof is trivial.
But the product is also a^(p-1).(p-1)! by commutativity. So a^(p-1)=1 (mod p).
There's a clever idea in there, to consider prod(ka) for fixed a and k=1..(p-1). Not finding the clever idea means you won't prove it. Finding the clever idea is tricky. Once you know the clever idea the proof is trivial.